pat 乙级 1012 部分测试点无法通过(C语言)

#测试点1、6无法通过

题目:

img

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以下是代码:

#include<stdio.h>
int main()
{
    int N;
    scanf("%d",&N);
    int s[N];
    for(int i=0;i<N;i++)
    scanf("%d",&s[i]);
    int a0[1000],a1[1000],a2[1000],a3[1000],a4[1000];
    int j=0,k=0,m=0,n=0,p=0;
    for(int i=0;i<N;i++)
    {
        if(s[i]%5==0)
        {
            a0[j]=s[i];
            j++;
        }
        else if(s[i]%5==1)
        {
            a1[k]=s[i];
            k++;
        }
        else if(s[i]%5==2)
        {
            a2[m]=s[i];
            m++;
        }
        else if(s[i]%5==3)
        {
            a3[n]=s[i];
            n++;
        }
        else
        {
            a4[p]=s[i];
            p++;
        }
    }
    //余数为0:
    if(j==0)
    printf("N ");
    else
    {
        int sum=0;
        for(int i=0;i<j;i++)
        {
            if(a0[i]%2==0)
            sum+=a0[i];
        }
        printf("%d ",sum);
    }
    //余数为1
    if(k==0)
    printf("N ");
    else
    {
        int sum=0;
        for(int i=0;i<k;i++)
        {
            if(i%2==0)
            sum=sum+a1[i];
            else
            sum=sum-a1[i];
        }
        printf("%d ",sum);
    }
    //余数为2
    if(m==0)
    printf("N ");
    else
    printf("%d ",m);
    //余数为3
    if(n==0)
    printf("N ");
    else
    {
        double average=0;
        for(int i=0;i<n;i++)
        average+=a3[i];
        average=1.0*average/n;
        printf("%.1f ",average);     //  %.nf表示保留n位小数
    }
    //余数为4
    if(p==0)
    printf("N");
    else
    {
        int max=a4[0];
        for(int i=0;i<p;i++)
        {
            if(max<a4[i])
            max=a4[i];
        }
        printf("%d",max);
    }
    return 0;
}

请求点拨

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你这个代码的算话实现的结构有些问题:
我给你我的代码吧:

#include <stdio.h>

int main() {
    int N;
    scanf("%d", &N);
    int arr[N], A[6] = {0}, cntA[6] = {0};
    for (int i = 0; i < N; i++) {
        scanf("%d", &arr[i]);
        if (arr[i] % 10 == 0) { //能被5整除的偶数
            A[1] += arr[i];
            cntA[1]++;
        } else if (arr[i] % 5 == 1) { //被5除后余1 
            cntA[2]++;
            if (cntA[2] % 2 == 1) { //第奇数个 
                A[2] += arr[i]; 
            } else {
                A[2] -= arr[i];
            }
        } else if (arr[i] % 5 == 2) { //被5除后余2
            A[3]++;
            cntA[3]++;
        } else if (arr[i] % 5 == 3) { //被5除后余3
            A[4] += arr[i];
            cntA[4]++;
        } else if (arr[i] % 5 == 4 && A[5] < arr[i]) { //被5除后余4
            A[5] = arr[i];
            cntA[5]++; 
        }
    }
    for (int i = 1; i <= 5; i++) {
        if (cntA[i] == 0) {
            if (i != 1)
                printf(" ");
            printf("N");
        } else if (i == 1) {
            printf("%d", A[1]);
        } else if (i == 2 || i == 3 || i == 5) {
            printf(" %d", A[i]);
        } else { // i = 4
            printf(" %.1f", (double)A[i] / cntA[4]);
        }
    } 
    printf("\n");
    return 0; 
}


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数字分类?

// PAT BasicLevel 1012
// https://pintia.cn/problem-sets/994805260223102976/problems/994805311146147840

#include <iostream>
using namespace std;

int main()
{
    // 数字个数
    int n;
    cin >> n;

    // 获取数字
    int* numArr=new int[n];
    for(int i=0;i<n;++i){
        cin >> numArr[i];
    }

    // 遍历数组,计算A1至A5
    int a1=0,a1Count=0;
    int a2=0,flag=1,a2Count=0;
    int a3=0,a3Count=0;
    double a4Sum=0,a4Count=0;
    int a5=0,a5Count=0;
    for(int i=0;i<n;++i){
        switch(numArr[i]%5){
        case 0:
            if(numArr[i]%2==0){
                a1+=numArr[i];
                a1Count++;
            }
            break;
        case 1:
            a2+=flag*numArr[i];
            flag=-flag;
            a2Count++;
            break;
        case 2:
            a3++;
            a3Count++;
            break;
        case 3:
            a4Sum+=numArr[i];
            a4Count++;
            break;
        case 4:
            if(numArr[i]>a5){
                a5=numArr[i];
                a5Count++;
            }
            break;
        }
    }

    // 输出A1至A5
    if(a1Count>0){
        cout << a1 << ' ';
    }else{
        cout <<"N ";
    }

    if(a2Count>0){
        cout << a2 << ' ';
    }else{
        cout << "N ";
    }

    if(a3Count>0){
        cout << a3 << ' ';
    }else{
        cout << "N ";
    }

    if(a4Count>0){
        printf("%.1lf ", a4Sum / a4Count);
    }else{
        cout << "N ";
    }

    if (a5Count > 0){
        cout << a5;
    }
    else{
        cout << 'N';
    }

    delete[] numArr;
    //system("pause");
    return 0;
}

用switch case